OG 11 DS Questions

Thats good. Focus on (1) DS traps that GMAT throws at you; and (2) focus on methods.

Lemme give you an example.

Q. Is x/(x+9) > 1/(x+1)

1. x < -3
2. x > 5

You know how to do these things: back solving or pick numbers. There is a better way to attack such questions if you dont assume that the question is not simplified enough for get a firm handle.

x/(x+9) - 1/(x+1) = (x-3)(x+3)/[(x+9)(x+1)]

or

Is (x+9)(x+3)(x+1)(x-5) > 0

1. x < -3
2. x > 5

B is sufficient.

palvarez wrote: (x-3)(x+3)/[(x+9)(x+1)]

or

Is (x+9)(x+3)(x+1)(x-5) > 0

1. x < -3
2. x > 5

B is sufficient.

Can you explain how you went from (x-3) (x+3) / (x+9)(x+1) to (x+9)(x+3)(x+1)(x-5)>0?

I do not see it.

Also, can I break down your statement like this:

x/(x+9) > 1/ (x+1)--> x/x+x/9>1/x+1/1
--> 1+x/9 > 1/x +1 -->
Is x/9 > 1/x? From here, it is easier to solve. Did I do that correct though?

chipbmk wrote:

palvarez wrote: (x-3)(x+3)/[(x+9)(x+1)]

or

Is (x+9)(x+3)(x+1)(x-5) > 0

1. x < -3
2. x > 5

B is sufficient.

Can you explain how you went from (x-3) (x+3) / (x+9)(x+1) to (x+9)(x+3)(x+1)(x-5)>0?

I do not see it.

Also, can I break down your statement like this:

x/(x+9) > 1/ (x+1)--> x/x+x/9>1/x+1/1
--> 1+x/9 > 1/x +1 -->
Is x/9 > 1/x? From here, it is easier to solve. Did I do that correct though?

it should be (x+9)(x+3)(x+1)(x-3). (x-5) was my mistake. I was tinkering with the problem to not have some weird numbers.

Well, you can always find a better way for a particular problem. I was focusing on a standard way, which is better than picking numbers.

x/(x+9) is not same as (x/x) + (x/9). Same with 1/(x+1).

(a+b)/x = (a/x) + (b/x)

x/(a+b) is not same as (x/a)+(x/b).


Whenever you see fractions in inequalities, there is a better way to deal with it.

Lemme give anotehr example.


Is x/y < 1.

(x/y) - 1 < 0
(x-y)/y < 0
(x-y)y/y^2 < 0
(x-y)y < 0, since y^2 is always +ve.

All these inequalities are same as long as y <> 0. But solving (x-y)y is easier than all other variants.

palvarez wrote:

chipbmk wrote:

palvarez wrote: (x-3)(x+3)/[(x+9)(x+1)]

or

Is (x+9)(x+3)(x+1)(x-5) > 0

1. x < -3
2. x > 5

B is sufficient.

Can you explain how you went from (x-3) (x+3) / (x+9)(x+1) to (x+9)(x+3)(x+1)(x-5)>0?

I do not see it.

Also, can I break down your statement like this:

x/(x+9) > 1/ (x+1)--> x/x+x/9>1/x+1/1
--> 1+x/9 > 1/x +1 -->
Is x/9 > 1/x? From here, it is easier to solve. Did I do that correct though?

it should be (x+9)(x+3)(x+1)(x-3). (x-5) was my mistake. I was tinkering with the problem to not have some weird numbers.

Well, you can always find a better way for a particular problem. I was focusing on a standard way, which is better than picking numbers.

x/(x+9) is not same as (x/x) + (x/9). Same with 1/(x+1).

(a+b)/x = (a/x) + (b/x)

x/(a+b) is not same as (x/a)+(x/b).


Whenever you see fractions in inequalities, there is a better way to deal with it.

Lemme give anotehr example.


Is x/y < 1.

(x/y) - 1 < 0
(x-y)/y < 0
(x-y)y/y^2 < 0
(x-y)y < 0, since y^2 is always +ve.

All these inequalities are same as long as y <> 0. But solving (x-y)y is easier than all other variants.

Thanks for the samples. That is good for me to focus on when I go back through all of the problems I got wrong. I am sure there will be at least a few that I can utilize what you are talking about.

Can you explain what "+ve" literally means. I have seen it and "-ve" used on this forum a lot, but dont know the actual meaning. I understand what is meant by it ... either positive or negative, but dont know what it literally means.

-2, -3, -1/5 -sqrt(2) are all negative. Any real number which is less than 0 are called negative.

2, 3, sqrt(5), 3/578578 are all positive, since they are greater than 0.

0 is neither positive, nor negative.

So, when one says that x is a non-negative integer, it means 0, 1, 2, 3, ...

non-negative = 0 and positive and vice versa.